# Homework Help: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this right?

1. Feb 3, 2013

### Lo.Lee.Ta.

1. Find the volume between y-10=x and y2 -6y =x, rotated around x=1.

2. R= y2 -6y -1

r= y-10 -1

∫2 to 5 of [($\pi$(y2 -6y -1)2) - $\pi$(y - 11)2]

∫$\pi$[y4 - 12y3 + 4y2 + 12y + 1] - $\pi$[y2 - 22y + 121]dy

= $\pi$[y5/5 - 12y4/4 + 4y3/3 +122/2 +y] - $\pi$[y3/3 - 22y2/2 + 121y] |2 to 5

...LONG SUBSTITUTION...

= -928.33$\pi$ - 371.67$\pi$ + 4.93$\pi$ + 200.67$\pi$

= -1094.4$\pi$

...Is this the right answer...? I don't really think it's right because it's negative!
But what am I doing wrong here? :(

Thanks SO much for your help! :D

2. Feb 3, 2013

### Dick

Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

It's easy to get something negative if you are subtracting something larger from something smaller when you meant to do the reverse. Which of those curves has the largest radius around the axis x=1? And I don't think just flipping the sign right will fix it either. Some other error around as well. Ah, you didn't square y^2-6y-1 correctly.

3. Feb 4, 2013

### Lo.Lee.Ta.

Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

You're right! I didn't square the (y2 - 6y -1) correctly! X(

The correct sqaring should equal: (y4 - 12y3 + 34y2 + 12y + 1)

So the integral seems like is should really be:

∫2 to 5 [$\pi$(y4 - 12y3 + 34y2 + 12y + 1) - $\pi$(y2 - 22y + 121)]dy

AWFUL AND LONG SUBSTITUTION!

= 321.67$\pi$ - 371.67$\pi$ - 75.07$\pi$ + 199.33$\pi$

= 74.26$\pi$

BUT when I put this whole integral into Wolfram Alpha (including the squares), it said it equals 0!

Then when I put in the integral again (this time with the squares factored out), it said the answer should be 15.9016!
What?! Would you please tell me what's wrong here?

Thank you SO much! :D

4. Feb 4, 2013

### Lo.Lee.Ta.

Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

I went through this thing again and again, and I only thing I found was a slight error making the answer 75.6$\pi$!

5. Feb 4, 2013

### Dick

Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

I think 75.6*pi sounds just fine. You KNOW 0 isn't correct, right? What exactly did you input to WA?

6. Feb 5, 2013

### Lo.Lee.Ta.

Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

integral from 2 to 5 of [pi(y^4 - 12y^3 + 34y^2 + 12y + 1) - pi(y^2 - 22y + 121)]

The above is exactly what I typed into WA.

7. Feb 5, 2013

### Dick

Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

That's an interesting misunderstanding. WA takes pi(x) to be a function giving the number of primes less than or equal to x. http://en.wikipedia.org/wiki/Prime-counting_function Try reading the fine print in the input box. If don't want that write pi*(x) instead.

Last edited: Feb 5, 2013